Energy, Work & Power (5 of 31) Gravitational Potential Energy, An Explanation

[@alanisquijano2031]

I just understood something that I've been trying to understand for hours in under 10 mins. Thank you! So helpful!

[@stepbystepscience]

Great that you got it straightened out. Thanks for the comment.

[@TimmehUrao]

I've been binge watching your videos sir and I don't even have to look at other videos to understand a topic, your videos/website is enough. Thank you so much!

[@stepbystepscience]

Great, thanks for watching and commenting! Did you subscribe?

[@TimmehUrao]

@@stepbystepscience Ofcourse sir!

[@gloryikuku8621]

Great class explained with sublime clarity.

[@stepbystepscience]

Many thanks, glad it was helpful!

[@jayantagupta3832]

Brilliant....i understood it just because of u....thank u very much😘😘

[@Isa7845]

Thank you ! Now all I need is to borrow your brain for a semester.

[@stepbystepscience]

Hmmmm...I will have to think about that....but the videos are always available. You can see a listing of all my videos at my website, www.stepbystepscience.com

[@pablovazquez9898]

This guy is a LEGEND!!!

[@stepbystepscience]

Nice of you to say, thanks.

[@ckimsey77]

Its probably negligible with a low mass object and great force source, but what about a higher mass object will the assumption of a=g with constant v still yield a tiny error? Error i refer to is the initial lifting, as an acceleration must exist for a small amt of time to begin moving the object upward...there's no v starting off. Also this boundary gradient time window could be more significantly large if the mass is large and force lower then constant v achieved during lift would take longer to reach, so the starting time where a is not zero would be larger...which would make actual value higher than estimate assuming constant v for all of t. The more i think about this, the more i confuse myself...would the slowing of the mass as it reaches desired height deacceleration cancel the added force to initially start the movement???? Or is this totally irrelevant because we're calculating net work, change in U, and thus being a change rather than absolute quantity (as chemical internal energy and majority of thermodynamics is a change in not absolute) the work is the same regardless how long it takes and/or independent of boundary exceptions to constant v assumption?? Ive always scratched my head reguarding all models/assumptions at what happens in the extremes where assumptions no longer work and the equation falls apart? Like d/dt (k * dT/dx) = 0 only accounting for 1-D heat flux and no heat gen (q dot=0) as well as þ*cp*(dT/dt) = 0 no heat energy stored, as i do not want to deal w the ridiculously long, complex, full length 2nd order partial differential equation thats longer than this text box....(how about deriving it w shell method? oh thats brutal 3 pages of diff equations; i love engineering...) anyway with these conditions the T gradient thru material is linear at steady state, and *in general* increases as thickness decreases (though wierdly at specific thicknesses insulators will actually suck heat out vs hold heat in, very odd.__). However, as material becomes very thin the equation shows flux increasing to infinitely high values which isnt possible; this bothers me... Are new models derived for such boundaries, maybe from quantum based principles because atomic interactions become more dominant at micron thicknesses? How do you know how thin is too thin; what determines the threshold where the flux eq. no longer works as written above??

[@youngun550]

Hi Step-by-Step Science. Really enjoying your videos. I've never had to square 9.8 before Is the final equation 1.5 x 96.04 x 0.45? This has really thrown me. Any help you can offer would be appreciated as the other method of 1.5 x 9.8 x 0.45 seems to have brought me to your answer. Just wanted a little clarification on why you mentioned 9.8m/s squared? Thanks and have a happy new year!

[@srinivasaraoyenugula8965]

Your website is very helpful Brian Swarthout

[@stepbystepscience]

+hashi yenugula Thank you very much!

[@dalilarobledodebasabe191]

Super well explained. Thank you for this.

[@stepbystepscience]

+Dalila Robledo De Basabe You are very welcome, You can see a listing of all my videos at my website, www.stepbystepscience.com

[@derricksajan1258]

well explained sir. good work

[@이완구-q7m]

Great presentation

[@stepbystepscience]

Thank you!

[@kawaeeee]

Perfect explanation!

[@stepbystepscience]

Tanks for watching and commenting!

[@shaleemmaqsood8551]

Good work you did. Helpful !

[@dulalghosh2071]

Sir if I throw a object up with a force F where F

[@stepbystepscience]

Those two forces are not really directly related. F(throw up) is the force you apply to something. F = mg is the weight of the object....need more info.

[@ptyptypty3]

if we lift a mass upward at a constant velocity... then we're doing work against gravity.. but since acceleration is the derivative of Velocity.. and velocity is Constant... then acceleration equals zero.. therefore the force equals zero.. and therefore the WORK must equal zero because work is equal to force times distance..... right?... and yet that's not right.. so how is that that is any different from if I were just to HOLD a mass up in the air.. with velocity equal to zero.. in that case I'm considered to NOT have done work..

[@stepbystepscience]

If the object is moving with a constant velocity the net force acting on the object is zero and the acceleration is zero. But the individual forces do not need to be zero. When you lift something up at a constant velocity, you are doing work on the object by applying a force against the force of gravity. If the object is not moving then there is not distance and no work is done. does that help?

[@priscillasue5228]

Perfect explanation. I

[@stepbystepscience]

Thank you, that is very nice!

[@rasheedahmad9077]

great video thank you so much!!

[@stepbystepscience]

You are very welcome. You can see a listing of all my videos at my website, www.stepbystepscience.com

[@prathameshrumde3027]

awwwwwwwwwwsom..........sir

[@wadjaafar]

Very Helpful. Thanks.

[@stepbystepscience]

Thanks for the comment. You can see a listing of all my videos at my website, www.stepbystepscience.com

[@arwaimad4749]

thanks your video always help me

[@stepbystepscience]

you are very welcome, thanks for commenting.

[@qiwenhuang5840]

Why is the gravitational acceleration here is positive for both cases?

[@stepbystepscience]

+Qiwen Huang Because potential energy is a scalar quantity. You can see a listing of all my videos at my website,www.stepbystepscience.com

[@آدم-ع9ي]

Legend of since

[@stepbystepscience]

Thanks.

[@mohameda.4713]

why do we have to move the object upward in a constant velocity ?

[@John-lf3xf]

Mohamed A. We dont

[@AragornHalt]

thank you friend ! :)

[@stepbystepscience]

You are very welcome, You can see a listing of all my vidoes at www.stepbystepscience.com

[@mrsrubyada9666]

tq very much. i already understand it!

[@stepbystepscience]

You are very welcome. You can see a listing of all my videos at my website, www.stepbystepscience.com

[@juzair6669]

i thought when force and d are perp the work is 0 ?

[@jhmkyt1855]

weener yum yum

[@rileykinner8986]

Thanks dude!

[@jhmkyt1855]

woah dude, what the frick is this

[@rileykinner8986]

@@jhmkyt1855 this video is helpfil to me study

[@jhmkyt1855]

@@rileykinner8986 still not going to help you pass the next test

[@rileykinner8986]

@@jhmkyt1855 yes i pass the next test idiot face

[@stepbystepscience]

You are very welcome, thanks for the comment.

[@pachie-information-scientist]

Master machinery

[@stepbystepscience]

Thanks

[@khushimehta9471]

Thank you very much. It was very helpful.

[@jospinlionel3423]

why -0,85 under the table?

[@stepbystepscience]

+jospin lionel Because you are moving the object down from its starting position. Up is positive and down is negative. You can see a listing of all my videos at my website, www.stepbystepscience.com

[@jospinlionel3423]

thank you sir

[@sagarbhatt-karia2403]

This was amazing

[@singjsatyam9513]

Why in y axis cos You taken

[@bluewisdom7000]

You did a mistake mgdelta y should be negative

[@stepbystepscience]

Really?

[@lanmisu]

too clear for just a physic knowledge neutral person like me to understand

[@theHaHa1091]

+mittar pwar lol